prove $\sum_{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$ prove  $\sum_\text{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$
I couldn't proceed much. I tried rearranging the inequality  and it became
$a^4c+b^4a+c^4b\ge a^2b^2c+b^2c^2a+c^2a^2b.$
I  tried using SOS here but it did not work.Also assuming $a\ge b\ge c$ didn't make things easier.
I also tried to work with one variable but it is a fourth degree so I skipped the calculus approach. We are actually supposed to prove using A-M G-M but other methods are also welcome.
 A: You can not assume that $a\geq b\geq c$ because the inequality is cyclic and not symmetric.
Since $(a^3,b^3,c^3)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ have an opposite ordering, by Rearrangement we obtain: $$\sum_{cyc}\frac{a^3}{b}\geq\sum_{cyc}\frac{a^3}{a}=\sum_{cyc}a^2\geq\sum_{cyc}ab,$$ where the last inequality it's also Rearrangement or it's just $$\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$
About Rearrangement see here: https://math.stackexchange.com/edit-tag-wiki/5774
In our case triples $(a^3,b^3,c^3)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ have an opposite ordering, which says
$$\sum_{cyc}\frac{a^3}{b}=\sum_{cyc}a^3\cdot\frac{1}{b}\geq \sum_{cyc}a^3\cdot\frac{1}{a}=\sum_{cyc}a^2.$$
Now, triples $(a,b,c)$ and $(a,b,c)$ have the same ordering, which says:
$$\sum_{cyc}a^2=\sum_{cyc}a\cdot a\geq\sum_{cyc}ab.$$
A: Also, SOS helps here:
$$\sum_{cyc}\left(\frac{a^3}{b}-ab\right)=\sum_{cyc }\frac{a(a^2-b^2)}{b}=$$
$$=\sum_{cyc}\left(\frac{a(a^2-b^2)}{b}-(a^2-b^2)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b}\geq0.$$
A: $a(a-b)^2\geq 0\implies a^3\geq 2a^2b-ab^2\implies \frac{a^3}{b}\geq2a^2-ab$ and so
$$\sum\frac{a^3}{b}\geq 2\sum a^2 - \sum ab\geq \sum ab.$$
Or you can directly use AM-GM by figuring out the coefficients by writing:
$$x\frac{a^3}{b} + y\frac{b^3}{c}+z\frac{c^3}{a}\geq (x+y+z)\sqrt[x+y+z]{a^{3x-z}b^{3y-x}c^{3z-y}} = ab,$$
for example.
A: Another way.
By AM-GM:
$$\sum_{cyc}\frac{a^3}{b}=\frac{1}{13}\sum_{cyc}\left(\frac{5a^3}{b}+\frac{6b^3}{c}+\frac{2c^3}{a}\right)\geq\sum_{cyc}\sqrt[13]{a^{15-2}b^{-5+18}c^{-6+6}}=\sum_{cyc}ab.$$
A: I wanted to present an elementary solution:
$$ {a^3 \over b} + ab \ge 2a^2 $$
and by $2$ other cyclic inequalities,
$$ \sum_{cyc}{a^3\over b}\geq 2(a^2+b^2+c^2)-ab-bc-ca\ge ab+bc+ac $$
Now it's sufficient,
$$ a^2+b^2+c^2\ge ab+bc+ca \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
